Maths in focus Chatper 3.9 ( Integration Volumes ): (1 Viewer)

Boonyak · Feb 19, 2012

Ex3.9)Q21.)
The are bounded by the curves [maths] y=x^2,y=(x+2)^2[/maths] is rotated about the x-axis. Find the volume of the solid of revolution found

SpiralFlex · Feb 19, 2012

- Draw a pretty diagram

Finding the points of intersection,

$x^2=x^2+4x+4$

$x=-1$

$y=1$

That's your point of intersection.

About the x axis,

$V=\pi \int_{-1}^{0}x^4 \; dx+\pi \int_{-2}^{-1}(x+2)^4 \; dx$

$=\pi [\frac{x^5}{5}]_{-1}^{0}+\pi [\frac{(x+2)^5}{5}]_{-2}^{-1}$

$=Spiral's favourite Number$

nightweaver066 · Feb 19, 2012

Diagram first.

First we want to find the points of intercepts.

So solving both of the equations,

$x^2 = (x + 2)^2$

$x^2 - (x + 2)^2 = 0$

$(2x + 2)(-2) = 0\;\;\;$I just did difference of two squares$$

$x = -1$

Now splitting the region in to two so we can work with them seperately.

First rotating the blue region around the x axis:

$V = \pi \int^{-1}_{-2} (x + 2)^4 dx$

$= \pi [\frac{(x + 2)^5}{5}]^{-1}_{-2}$

$= \pi ( \frac{1}{5} - 0)$

$= \frac{\pi}{5}units^3$

Now for the red region rotated around the x-axis:

$V = \pi \int^{0}_{-1} x^4 dx$

$= \pi [\frac{x^5}{5}]^{0}_{-1}$

$= \pi (0 - ( - \frac{1}{5}))$

$= \frac{\pi}{5}units^3$

Now adding these two together,

$\frac{\pi}{5} + \frac{\pi}{5} = \frac{2\pi}{5}units^3$

Boonyak · Feb 19, 2012

Thanks i get it know

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Maths in focus Chatper 3.9 ( Integration Volumes ): (1 Viewer)

Boonyak

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